POJ 3463 Sightseeing

求最短路，和最短路长度加1的路的条数。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;
const int N = 1009;
const int M = 10009;
const int INF =0x3f3f3f3f;
int n,m,dis[N<<1],cnt[N<<1],ST,EN;
bool visit[N<<1];
struct node
{
    int to,nex,dis;
} L[M];
int F[N],cntF;
void add(int f,int t,int dis)
{
    L[cntF].dis = dis;
    L[cntF].nex = F[f];
    L[cntF].to = t;
    F[f] = cntF++;
}
void init()
{
    scanf("%d%d",&n,&m);
    memset(cnt,0,sizeof(cnt));
    memset(F,0,sizeof(F));
    cntF = 1;
    int f,t,dis;
    for(int i=0; i<m; i++)
    {
        scanf("%d%d%d",&f,&t,&dis);
        add(f,t,dis);
    }
    scanf("%d%d",&ST,&EN);
}
struct nod
{
    int dis,to;
    bool operator<(const nod t) const
    {
        return dis>t.dis;
    }
};
priority_queue<nod> que;
void solve()
{
    while(!que.empty()) que.pop();
    memset(visit,false,sizeof(visit));
    memset(dis,INF,sizeof(dis));
    nod e,t;
    e.to = ST;
    e.dis = 0;
    que.push(e);
    dis[ST] = 0;
    cnt[ST] = 1;
    cnt[ST+N] = 1;
    while(!que.empty())
    {
        e =que.top();
        que.pop();
        if(visit[e.to]) continue;
        visit[e.to] = true;

        for(int i=F[(e.to>N?e.to-N:e.to)]; i; i=L[i].nex)
        {
            int to = L[i].to;
            int tmp=dis[e.to]+L[i].dis;
            if(dis[to]>tmp)
            {
                if(dis[to]!=INF)
                {
                    dis[to+N] = dis[to];
                    cnt[to+N] = cnt[to];
                    t.dis=dis[to+N];
                    t.to=to+N;
                    que.push(t);
                }
                dis[to] = tmp;
                cnt[to] = cnt[e.to];
                t.dis = dis[to];
                t.to = to;
                que.push(t);
            }
            else if(dis[to] == tmp)
            {
                cnt[to] += cnt[e.to];
            }
            else if(dis[to+N] > tmp)
            {
                dis[to+N] = tmp;
                cnt[to+N] = cnt[e.to];
                t.dis = dis[to+N];
                t.to = to+N;
                que.push(t);
            }
            else if(dis[to+N] == tmp)
            {
                cnt[to+N] += cnt[e.to];
            }
        }
    }
    printf("%d\n",cnt[EN]+(dis[EN+N]-dis[EN]==1?cnt[EN+N]:0));
}
int main()
{
    freopen("in.txt","r",stdin);
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        init();
        solve();
    }
    return 0;
}